\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)

a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)

\(=\dfrac{2}{3}+\dfrac{2}{7}\)

\(=\dfrac{14}{21}+\dfrac{6}{21}\)

\(=\dfrac{20}{21}\)

\(\dfrac{7}{12}-\dfrac{27}{7}.\dfrac{1}{8}=\dfrac{7}{12}-\dfrac{27}{56}=\dfrac{392}{672}-\dfrac{324}{672}=\dfrac{68}{672}=\dfrac{17}{168}\)

a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)

b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)

c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)

d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)

Mong bạn xem lại đề bài.

\(49,=\dfrac{38}{5}-\left(\dfrac{19}{7}+\dfrac{28}{5}\right)\)

\(=\dfrac{38}{5}-\dfrac{19}{7}-\dfrac{28}{5}\)

\(=\left(\dfrac{38}{5}-\dfrac{28}{5}\right)-\dfrac{19}{7}\)

\(=2-\dfrac{19}{7}=-\dfrac{5}{7}\)

\(50,=\dfrac{25}{81}.\dfrac{15}{22}=\dfrac{125}{594}\)

Câu 50 tus sửa đề 

\(-\dfrac{1}{9}.\dfrac{15}{22}:\dfrac{-25}{9}=-\dfrac{1}{9}.\dfrac{15}{22}.\dfrac{-9}{25}=\dfrac{15}{22}.\dfrac{1}{25}=\dfrac{3}{110}\)

Bài 1 :

Câu 49 . = 38/5 - ( 19/7 + 28/5 )

               = 38/5 - 19/7 - 28/5 

               = ( 38/5 - 28/5 ) - 19/7 

               = 2 - 19/7 = -5/7

Câu 50 . = ( -1/9 : -25/9 ) . 15/22

               = 1/25 . 15/22 = 3/110 

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Bạn Tiểu Tôm Béo ơi có thể check lại đề không ạ?

Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)

Cộng vế với vế ta được 

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{20^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(\Rightarrow T< 2-\dfrac{1}{20}=\dfrac{39}{20}\)

mà 39/20 < 8/7 => T < 8/7 

\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)

=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)

=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)

=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)

=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)

=\(\dfrac{-79}{16}\)

\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)

=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)

=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)

=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)

\(\dfrac{4}{7}.\dfrac{89}{5}-\dfrac{4}{5}.\dfrac{82}{7}\)

=\(\dfrac{4}{5}.\dfrac{89}{7}\dfrac{4}{5}.\dfrac{82}{7}\)

=\(\dfrac{4}{5}\left(\dfrac{89}{7}-\dfrac{82}{7}\right)\)

=\(\dfrac{4}{5}.1=\dfrac{4}{5}\)

\(\approx\) chúc bạn học tốt \(\approx\)